使用 JAXB 将 Java 列表转换为 XML 编码不起作用？

Question

在发布此问题之前，我对类似类型问题的堆栈交换答案进行了大量研究。我也尝试了很多解决方案来解决谷歌的问题。最后我想我可以在这里问问题:

1. 我正在尝试使用 JAXB 编码过程将 Java 帐户对象列表转换为 XML。但预期的是下面的前 15 行正确的 xml，当前生成的是最后 2 行代码:

这里是:

 <?xml version="1.0" encoding="UTF-8" standalone="yes"?>
  <accounts>
    <account>
      <accountName>Media Team </accountName>
      <ownerName>Support Team</ownerName>
      <ownerEmail>sbjba@gmail.com</ownerEmail>
      <ownerId>1221323</ownerId>
    <account>
    <account>
       <accountName>Delivery</accountName>
       <ownerName>jadajsnfd</ownerName>
       <ownerEmail>bfjaja@gmail.com</ownerEmail>
       <ownerId>82487282</ownerId>
   </account>
 </accounts>

 <?xml version="1.0" encoding="UTF-8" standalone="yes"?>
 </accounts>

这是 Accounts 类:

@XmlAccessorType(XmlAccessType.FIELD)
@XmlRootElement(name = "accounts")
public class Accounts {
    @XmlElement(name = "account", type = Account.class)
    private List<Account> accounts = new ArrayList<Account>();
public Accounts(){
  }
public Accounts(List<Account> accounts) {
 }
public List<Account> getAccounts() {
    return accounts;
 }
public void setAccounts(List<Account> accounts) {
    this.accounts = accounts;
 }
}

这是帐户类:

@XmlAccessorType(XmlAccessType.FIELD)
@XmlRootElement(name = "account")
public class Account {
    String accountName;
    String ownerName;
    String ownerEmail;
    String ownerId;
//All setters and getters are here public
}

最后这是我的代码:

File videosFile = new File("C:/AccountInfo.xml");
List<Account> accounts = null;
List<KalturaMediaEntry> mediaList = null;
mediaList = getAllLatestMedia();
    if (mediaList.size() >= 1) {
        System.out.println("mediaList.size() ---------->"
                + mediaList.size());
        accounts = setMediaDataByAccount(mediaList);
        if (accounts.size() >= 1) {
        System.out.println("videos.size() --------->"
                + accounts.size());
        marshalVideos(accounts, videosFile);
        }
    }

public static void marshalVideos(List<Account> accounts,
        File videosFile) throws JAXBException, IOException {
    BufferedWriter writer = new BufferedWriter(new OutputStreamWriter(
            new FileOutputStream(videosFile, true), "UTF-8"));
    final Marshaller m = JAXBContext.newInstance(Accounts.class)
            .createMarshaller();
    m.setProperty(Marshaller.JAXB_FORMATTED_OUTPUT, true);
    m.marshal(new Accounts(accounts), writer);
    writer.close();
}

我也尝试过添加 @XMLSeeAlso 注释等解决方案。但不工作。请帮我解决这个问题。

提前致谢，拉吉

Answer 1

在 Accounts 的构造函数中，您不使用 Account 列表:

public Accounts(List<Account> accounts) {
}

所以，将其更改为

public Accounts(List<Account> accounts) {
    this.accounts = accounts;
}